1: Ta có: \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
\(=x^6\left(x-2\right)^2\left(x+2\right)^2\)
2: Ta có: \(m^3+27\)
\(=\left(m+3\right)\left(m^2-3m+9\right)\)
3: Ta có: \(x^3+8\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
4: Ta có: \(\frac{1}{27}+a^3\)
\(=\left(\frac{1}{3}+a\right)\left(\frac{1}{9}-\frac{a}{3}+a^2\right)\)
5: Ta có: \(8x^3+27y^3\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6: Ta có: \(\frac{1}{8}x^3+8y^3\)
\(=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
7: Ta có: \(8x^6-27y^3\)
\(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
8: Ta có: \(\frac{1}{8}x^3-8\)
\(=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
9: Ta có: \(\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)
10: Ta có: \(\left(a+b\right)^3-c^3\)
\(=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)\cdot c+c^2\right]\)
\(=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)
11: Ta có: \(x^3-\left(y-1\right)^3\)
\(=\left[x-\left(y-1\right)\right]\cdot\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]\)
\(=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
12: Ta có: \(x^6+1\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
1) \(x^{10}-4x^8+4x^6\)
\(=x^6\left(x^4-4x^2+4\right)\)
2) \(m^3+27=m^3+3^3=\left(m+3\right)\left(m^2-3m+3^2\right)\)
3) \(x^3+8=x^3+2^3=\left(x+2\right)\left(x^2-2x+2^2\right)\)
4) \(\frac{1}{27}+a^3=\left(\frac{1}{3}\right)^3+a^3=\left(\frac{1}{3}+a\right)\left[\left(\frac{1}{3}\right)^2-\frac{1}{3}a+a^2\right]\)
5) \(8x^3+27y^3=\left(2x\right)^3+\left(3y\right)^3=\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
6) \(\frac{1}{8}x^3+8y^3=\left(\frac{1}{2}x\right)^3+\left(2y\right)^3=\left(\frac{1}{2}x+2y\right)\left[\left(\frac{1}{2}x\right)^2-\frac{1}{2}x.2y+\left(2y\right)^2\right]=\left(\frac{1}{2}x+2y\right)\left(\frac{1}{4}x^2-xy+4y^2\right)\)
8) \(\frac{1}{8}x^3-8=\left(\frac{1}{2}x\right)^3-2^3=\left(\frac{1}{2}x-2\right)\left[\left(\frac{1}{2}x\right)^2+\frac{1}{2}x.2+2^2\right]=\left(\frac{1}{2}x-2\right)\left(\frac{1}{4}x^2+x+4\right)\)
10) \(\left(a+b\right)^3-c^3=\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]=\left(a+b-c\right)\left[\left(a^2+2ab+b^2\right)+ac+bc+c^2\right]=\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)11) \(x^3-\left(y-1\right)^3=\left(x-y+1\right)\left[x^2+x\left(y-1\right)+\left(y-1\right)^2\right]=\left(x-y+1\right)\left[x^2+xy-x+\left(y^2-2y+1\right)\right]=\left(x-y+1\right)\left(x^2+xy-x+y^2-2y+1\right)\)
P/s: Đăng ít thôi chớ bạn!
